The energy given to the lattice is much reduced, but not zero.
In a two-body interaction, if you push the bodies apart the ratio of the masses corresponds with the change in KE of the two bodies. The more mass one of the objects has, the less energy the interaction gives to it (even though the momentum change is identical).
In the nuclear decay, one of the "objects" is a gamma ray. If the other is an atomic nucleus, then it can get a significant fraction of the energy. If however you are able to treat the crystal lattice as the other object, the much greater mass means that almost all the energy goes into the gamma ray.
The lattice still receives the entire impulse or change in momentum due to the interaction. Just that the energy associated with it is tiny.
I didn't recognize that the big portion of your question was about the phonons.
the theoretical energy of the recoil is very, very small, and is quantified into phonons that have a minimal energy.
Almost. The energy of the recoil can be partitioned into vibrational energy within the lattice and bulk motion of the entire lattice. Normally the energy term is dominated by these vibrations, which can be quantized into phonons. In certain conditions, the vibrational energy is too low and is excluded, so the vibrational energy/phonons is exactly zero. But the entire lattice still receives momentum from the interaction. Because the mass of the lattice can be quite large, this will drive the change in velocity (and therefore the actual energy received) down to very low levels, but not theoretically zero.
...phonons take vibrational energy that is inherently different to energy associated with momentum change, and all the wikipedia fuzz about phonons is not to lose /another/ part of the energy to vibrations!
Rather than say that the vibrational energy is inherently "different", I would tend to think of it more like another "container" for the energy.
If that container is excluded at the time of the interaction, the the only locations for it are the gamma ray and the KE of the entire lattice. If that container is available, then it can hold some of the energy and the the range for the gamma is much less constricted.
